# Ok so this is a follow up question to my previous one: http://www.enotes.com/homework-help/have-solution-kno3-690g-1-6l-add-50g-k2cro4-463166#answer-644784 because the answer didn´t fully answer...

Ok so this is a follow up question to my previous one: http://www.enotes.com/homework-help/have-solution-kno3-690g-1-6l-add-50g-k2cro4-463166#answer-644784

because the answer didn´t fully answer me how I can find out at which temperatur the precipitate will form. Actually I have an idea that if I find deltaH and deltaS, then I can calculate deltaG* and if I get a positive number then the reaction is not spontanious and it´s not occuring, precipitate will form?

But I know KNO3 precipitate will form by itself, but I have to find out at which temperature it´ll form if K2CrO4 is added in the solution, and I hypothethised that it´ll be at a higher temperatur than without. Because as a solution of just KNO3 it´ll cool down and precipitate will form, but with K2CrO4 in the solution the sollubility is decreasing and precipitate can form at a much higher temperatur?

Actually I do have a graph for the reaction of KNO3 alone and I did find deltaH and deltaS as I did use in the calculations and added in the additional picture in the last question.

But I´m not sure how I can read from the graph at which temperatur precipitate will form? I´ve calculated at which temperature deltaG=0, and I got it´s 297,60K does it have something to do with the question earlier?

But I also calculated another InQ as when K2CrO4 is added to solution, do I need to add it to the graph too, in order to find out the new temp at which precipitate will form?

Also another info from my prelab is the volume did not change as I add 50g K2CrO4 to the solution. But I hope somebody can help me with this, I´m so confused right now and I couldn´t find much when I tried to browse the net either.

*print*Print*list*Cite

### 3 Answers

Let the precipitation of solid KNO3 starts at T K.

From the plot of lnK vs 1/T,

`-(DeltaH^o)/R=-4504.4`

`rArr DeltaH^o=4504.4*8.314=37449.6` J/mol

And, `(DeltaS^o)/R=17.552`

`rArr DeltaS^o=17.552*8.314=145.7` J/K-mol

Again, considereing the process of dissolution,

`K=6.83*6.83=46.7`

`lnK=ln46.7=3.843`

Assuming the value of `DeltaH^o` and `DeltaS^o` remain constant in the given temperature range,

`DeltaG^o at T K=DeltaH^o-T*DeltaS^o`

`=37449.6-145.7*T`

Again, `DeltaG^o=-RTlnK`

`=-8.314*T*3.843`

`=-31.9507T`

Equating the two expressions of `DeltaG^o,`

`-31.9507T=37449.6-145.7*T`

`rArr T(145.7-31.9507)=37449.6`

`rArr T=37449.6/(145.7-31.9507)`

`=329.2K`

`=56.1^oC`

-------------------------------------------

When K2Cr2O7 is added, it dissociates as:

`K2Cr2O7 stackrel rarr larr 2K^+ + Cr_2O_7^-`

Such that `[K^+]=6.83+(2*50)/(194.2*1.6)`

=7.15 M

`[NO_3^-]` =6.83 M

K'=7.15*6.83=48.83

lnK'=3.888

Calculating as above, the temperature when precipitation just begins now, T’ is `57.1^oC`

The results are in line with your expectation that addition of common ion (`K^+` ) raises the temperature at which precipitation of KNO3 begins by close to a degree.

I notice I had an calculation error with deltaS and I fixed it:

But now that I calculate T at deltaG = 0, I got -1837,67K and this is weird. Did I do something wrong?

**Images:**

Additional picture:

**Images:**